# Surface integral ex2 part 2: Evaluating integral | Multivariable Calculus | Khan Academy

Now that we've set up the parametrization,
let's try to evaluate the integral.
And the next thing we'll do is essentially
try to express ds right over here in terms of du and dv.
And we've seen this before.
ds is going to be equal to the magnitude of the cross product
of the partial of r with respect to u crossed
with the partial of r with respect to v du dv.
So first, let's take the cross product,
and we'll do that with a 3 by 3 matrix.
I'll do that right over here.
We set up, and I'll just kind of fill in what r sub u and r
sub v is in the actual determinant right over here.
So first we have our components, i, j, and k.
And now first, let's think about what
r sub u is, the partial of r with respect to u.
Well, its i component is going to be 1.
The partial of u with respect to u is just 1,
so its i component is 1.
Its j component is going to be 0,
partial of v with respect to u is
0. v does not change with respect to u,
so this is going to be 0.
This should be parentheses around here.
Partial of this with respect to u is just going to be 1 again.
The partial of v squared with respect to u is just 0.
So this is just 1 again.
And then r sub v, the partial of r with respect to v,
the i component is going to be 0,
j component is going to be 1, and the partial
of u plus v squared with respect to v is going to be 2v.
So it's a pretty straightforward determinant,
so let's try to evaluate it.
So the i component, it's going to be i.
So we cross out this column, this row.
It's going to be 0 times 2v minus 1 times 1.
So essentially, it's just going to be negative 1 times i.
So we're going to have negative i,
so this is equal to negative i.
And then the j component-- and we're
going to have to put a negative out front, because, remember,
we do that checkerboard pattern.
So cross out that row, that column.
1 times 2v is 2v.
Let me make sure I got that.
1 times 2v is 2v minus 0 times 1.
So it's just going to be 2v.
But since it was the j component, which
is going to be negative, it's going to be negative 2vj.
Let me make sure I did that right.
That column, that row, 1 times 2v is 2v minus 0 is 2v.
The checkerboard pattern, you'd have a negative j.
So you have negative 2vj.
And then we have find the k component.
Cross out that row, that column, 1 times 1 minus 0.
So it's going to be plus k.
So if we want the magnitude of this,
this whole thing right over here is just
going to be the square root-- I'm just
taking the magnitude of this part right
over here, the actual cross product.
It's going to be negative 1 squared, which
is just 1, plus negative 2v squared, which
is 4v squared, plus 1 squared, which is just 1.
So this whole thing is going to evaluate
to-- we have 2 plus 2v squared du dv.
Or actually, let me-- I almost made a mistake.
That would have been a disaster.
2 plus 4v squared du dv.
And if we want, maybe it'll help us
a little bit if we factor out a 2 right over here.
This is the same thing as 2 times 1 plus 2v squared du dv.
If you factor out the 2, you get this
is equal to the square root of 2 times the square root of 1
plus 2v squared du dv.
And I now think we are ready to evaluate the surface integral.
So let's do it.
All right.
So let me just write this thing down here.
So I'm going to write everything that has to do with v,
I'm going to write in purple.
So I'm just going to write the ds part right over here.
It is the square root of 2 times the square root of 1
plus 2v squared.
And then we're going to have du, which
I'll write in green, du, and then dv.
So this is just the ds part.
This is just the ds.
Now, we have the y right over here.
y is just equal to v. All right, so I'll write that in purple.
So y is just equal to v. That is y-- let me make it very clear--
and all of this is ds.
And now I can write the bounds in terms of u
and v. And so the u part, u is the same thing as x.
It goes between 0 and 1.
And then v, v is the same thing as y,
and y goes between, or v goes between, 0 and 2.
And I now think we're ready to evaluate.
And the u and v variables are not so mixed up,
so we can actually separate out these two integrals,
make this a product of two single integrals.
The first thing, if we look at it with respect to du,
all this stuff in purple is just a constant with respect to u,
so we can take it out of the du integral.
We can take all this purple stuff out of this du integral.
And so this double integral simplifies to the integral
from 0 to 2 of-- I'll write it as the square root of 2v times
the square root of 1 plus 2v squared.
So I factored out all of this stuff.
And then you have times the integral from 0 to 1 du,
and then times d, and then you have dv.
And now, if this was really complicated, I could say,
OK, this is just going to be a function of u.
It's a constant with respect to v.
And you could factor this whole thing out
and separate the integrals.
But this is even easier.
This integral is just going to evaluate to 1.
So this whole thing just evaluates to 1.
And so we've simplified this into one single integral.
So this simplifies to-- and I could even
take the square root of 2 out front-- the square root
of 2 times the integral from v is equal to 0 to v
is equal to 2 of v times the square root
of 1 plus 2v squared dv.
And so we really are in the home stretch
of evaluating this surface integral.
And here, this is basic.
This is actually a little bit of u substitution
that we can do in our head.
If you have a function, or kind of this embedded function
right over here, 1 plus 2v squared,
what's the derivative of 1 plus 2v squared?
Well, it would be 4v.
And we have something almost 4v here.
We can make this 4v by multiplying it by 4 here,
and then dividing it by 4 out here.
This doesn't change the value of the integral.
And so now this part right over here,
it's pretty straightforward to take the antiderivative.
The antiderivative of this is going
to be-- we have this embedded function's derivative right
over here, so we can kind of just treat it like an x or a v.
Take the antiderivative with respect
to this thing right over here, and we
get this is essentially 1 plus 2v squared to the 1/2 power.
We increment it by 1, so it's 1 plus 2v
squared to the 3/2 power.
And then you divide by 3/2, or you multiply by 2/3,
so times 2/3.
So this is the antiderivative of that.
And then, of course, you still have all of this stuff
out front-- square root of 2 over 4
And we are going to evaluate this from 0 to 2.
And actually, just to simplify it, let me factor out the 2/3.
We don't have to worry about that at 2 and 0.
So I'm going to factor out the 2/3, so times 2 over 3.
And actually, this will cancel out.
That becomes a 1.
That becomes a 2.
And so we are left-- this stuff over here is 1/6 times.
And now if you evaluate this at 2, you have 2v squared.
That's going to be 2 times 4 is 8 plus 1
is 9, 9 to the 3/2 power.
So 9 to the 1/2 is 3, 3 to the 3rd is 27.
So it's going to be equal to 27.
And then minus this thing evaluated at 0.
Well, this evaluated at 0 is just going to be 1.
1 to the 3/2 is just 1, so minus 1.
And so this gives us-- oh, I almost made a mistake.
There should be a square root of 2 up here,
square root of 2 over 6 times 27 minus 1.
So drum roll.
This gives us the value of our surface integral is,
let's see, 27 minus 1 is 26.
So we get 26 times the square root of 2 over 6.
And we can simplify it a little bit more.
Divide the numerator and denominator by 2,
and you get 13 square roots of 2 over 3.
And we are done.